将 xml 文件的条目转换为 Pandas 数据框

Question

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我有以下 xml 文件。 我想提取所有以 ItemDescription 开头的行并从中创建一个数据框，这样一列包含颜色，另一列包含 ID，另一列包含字母等等。 我怎样才能做到这一点？

我使用xml.etree.ElementTree包进行了尝试，但我无法生成数据xml.etree.ElementTree ，因为我无法访问所需行中的元素。 我不想使用pandas_read_xml因为它只适用于 pip，我猜。 即使我更新了熊猫， pd.read_xml也不起作用。 是否有一种可靠的方法可以使用xml.etree.ElementTree或其他不太花哨的包来实现？

<?xml version="1.0" ?>
<OrderList>
   <ItemDescriptions>
      <ItemDescription Color="rosybrown" ID="0" Letter="a" Type="Letter" Weight="1.67"/>
      <ItemDescription Color="lightcoral" ID="1" Letter="a" Type="Letter" Weight="0.91"/>
      <ItemDescription Color="indiaread" ID="2" Letter="a" Type="Letter" Weight="0.62"/>
      <ItemDescription Color="brown" ID="3" Letter="a" Type="Letter" Weight="2.92"/>
      <ItemDescription Color="firedbrick" ID="4" Letter="a" Type="Letter" Weight="2.34"/>
      <ItemDescription Color="maroon" ID="5" Letter="a" Type="Letter" Weight="0.53"/>
      <ItemDescription Color="darkred" ID="6" Letter="a" Type="Letter" Weight="2.72"/>
   </ItemDescriptions>
   <ItemBundles/>
   <Orders>
      <Order TimeStamp="">
         <Positions>
            <Position Count="1" ItemDescriptionID="9"/>
            <Position Count="1" ItemDescriptionID="18"/>
         </Positions>
      </Order>
      <Order TimeStamp="">
         <Positions>
            <Position Count="2" ItemDescriptionID="9"/>
            <Position Count="1" ItemDescriptionID="12"/>
            <Position Count="2" ItemDescriptionID="14"/>
            <Position Count="1" ItemDescriptionID="18"/>
            <Position Count="1" ItemDescriptionID="16"/>
         </Positions>
      </Order>
   </Orders>
</OrderList>

Answer 1

使用read_xml和xpath ：

>>> pd.read_xml('data.xml', xpath='./ItemDescriptions/ItemDescription')

        Color  ID Letter    Type  Weight
0   rosybrown   0      a  Letter    1.67
1  lightcoral   1      a  Letter    0.91
2   indiaread   2      a  Letter    0.62
3       brown   3      a  Letter    2.92
4  firedbrick   4      a  Letter    2.34
5      maroon   5      a  Letter    0.53
6     darkred   6      a  Letter    2.72

替代lxml ：

from lxml import etree

tree = etree.parse('data.xml')

df = pd.DataFrame([dict(elmt.items()) 
                       for elmt in tree.xpath('.//ItemDescription')])

Answer 2

使用 ElementTree（不需要外部库）

import xml.etree.ElementTree as ET
import pandas as pd

xml = '''<?xml version="1.0" ?>
<OrderList>
   <ItemDescriptions>
      <ItemDescription Color="rosybrown" ID="0" Letter="a" Type="Letter" Weight="1.67"/>
      <ItemDescription Color="lightcoral" ID="1" Letter="a" Type="Letter" Weight="0.91"/>
      <ItemDescription Color="indiaread" ID="2" Letter="a" Type="Letter" Weight="0.62"/>
      <ItemDescription Color="brown" ID="3" Letter="a" Type="Letter" Weight="2.92"/>
      <ItemDescription Color="firedbrick" ID="4" Letter="a" Type="Letter" Weight="2.34"/>
      <ItemDescription Color="maroon" ID="5" Letter="a" Type="Letter" Weight="0.53"/>
      <ItemDescription Color="darkred" ID="6" Letter="a" Type="Letter" Weight="2.72"/>
   </ItemDescriptions>
   <ItemBundles/>
   <Orders>
      <Order TimeStamp="">
         <Positions>
            <Position Count="1" ItemDescriptionID="9"/>
            <Position Count="1" ItemDescriptionID="18"/>
         </Positions>
      </Order>
      <Order TimeStamp="">
         <Positions>
            <Position Count="2" ItemDescriptionID="9"/>
            <Position Count="1" ItemDescriptionID="12"/>
            <Position Count="2" ItemDescriptionID="14"/>
            <Position Count="1" ItemDescriptionID="18"/>
            <Position Count="1" ItemDescriptionID="16"/>
         </Positions>
      </Order>
   </Orders>
</OrderList>'''

root = ET.fromstring(xml)
data = [desc.attrib for desc in root.findall('.//ItemDescription')]
df = pd.DataFrame(data)
print(df)

输出

        Color ID Letter    Type Weight
0   rosybrown  0      a  Letter   1.67
1  lightcoral  1      a  Letter   0.91
2   indiaread  2      a  Letter   0.62
3       brown  3      a  Letter   2.92
4  firedbrick  4      a  Letter   2.34
5      maroon  5      a  Letter   0.53
6     darkred  6      a  Letter   2.72